Chapter 3ContentsAtomic and Molecular WeightsThe MoleMolar massEMPIRICAL FORMULACOMBUSTION ANALYSISMOLECULAR FORMULASTOICHIOMETRYSTOICHIOMETRY2LIMITING REAGENTSYIELDS IN CHEMICAL REACTIONSSummaryJohn A. SchreifelsChem 211 1Chapter 3Calculations involving Chemical Formulae and EquationsJohn A. SchreifelsChem 211 2Contents•Mass and Moles of a Substance–Molecular weight–Moles•Determining Chemical Formulae–Mass % from formula–Elemental Analysis: gives % C, H, O–Determining formula from elemental analysis•Stoichiometry–Amounts of substances in a reaction–Limiting reagentsJohn A. SchreifelsChem 211 3Atomic and Molecular Weights•Molecular Weight (Formula weight): Summed masses of all atoms in the formula unit of the compound.•How?•For the general formula AaBbCcDd... fm = formula mass = aMA + bMB + cMC + dMD + ...•E.g. determine the formula mass (formula weight) of the following:•NaOH, H2O, CaCO3, C2H6O•Percentage composition from formula (mass % contribution of each element to the total mass of the compound). For AaBbCc•E.g. determine the mass % composition of each element in: NaOH, H2O, CaCO3, C2H6O100FWmaA%A-John A. SchreifelsChem 211 4The Mole•Since individual atoms are very small, they weigh very little. It is not convenient deal with masses that small, since we can typically measure gram amounts in the lab.•Mole: Number of atoms (or molecules) as there are in 12.00 g Carbon-12; 1 mol = 6.02x1023 and is called Avogadro's number.•Mole is used to indicate a certain number of particles (like a dozen or bushel might be used).1 mol of •water occupies approximately 18 mL and has 6.02x1023 molecules. •gold occupies approximately 10 mL and has 6.02x1023 atoms.John A. SchreifelsChem 211 5Molar mass•1 mol = FM = 6.02x1023 particles•The definition of a mol allows us to perform a variety of conversions. •Let n = # mol, n’ = # atoms (molecules), m = mass and FM = formula mass then we can convert to: •m from n, FME.g. Determine the mass of NaCl needed to have 5.0 mol of it.•n from m, FME.g. Determine the number of mol of NaCl in 15.00 g of it.•n’ from nE.g. Determine the number of molecules in 3.222 mol of NaCl•n’ from m, FME.g. Determine the number of atoms in 4.32 g of NaCl•m from n’, FME.g. Determine the mass of one formula unit of NaCl.John A. SchreifelsChem 211 6EMPIRICAL FORMULAEmpirical Formula: Simplest formula where all coefficients are integers.•In earlier example Fe2O3, Fe4O6 Fe6O9 Fe8O12 possible formulas for iron oxide. Its empirical formula Fe2O3. •Often obtained from % composition data.–Assume 100 g of compound- –Determine mol of each element.–Divide by smallest # of moles.–Multiply by smallest whole number to make all coefficients whole numbers.•E.g. Determine the empirical formula of a compound with the following % compositions:Mass%O = 34.7%Mass%C = 52.1%Mass%H = 13.1%John A. SchreifelsChem 211 7COMBUSTION ANALYSIS•Experimental mass % also determined. For organic compounds the sample is combusted. Analyzes the amount of C,H, and O in compounds.–C converted to CO2 and mass measured.–H converted to H2O and mass measured•E.g.: Combustion analysis of a 1.621 g sample of ethanol, which contains only C,H,O, was performed. Calculate the mass % composition of each element in ethanol.•Results: Mass CO2 = 3.095 gMass of H2O = 1.902 g.John A. SchreifelsChem 211 8MOLECULAR FORMULA•Measure molecular mass •Multiply empirical mass by integer to obtain molecular mass.•Multiply all the co-efficients by this integer.•E.g. Determine the molecular mass of a compound with the empirical formula NO2 and a molecular mass of 92.00 g/mol.John A. SchreifelsChem 211 9STOICHIOMETRYThe relative amounts of reactants and products in a reaction are given by the ratio of stoichiometric coefficients. (Conservation of mass).•E.g. For the reaction : •2Na(s) + Cl2(g)  2NaCl(s)•2 mol Na = 1 mol of Cl2 = 2 mol of NaCl.•E.g. 2 Determine # mol of Cl2 needed to react with 4.2 mol of Na. How many mol of NaCl will form?•Mol of Cl2•Mol NaCl:•Summary: For aA + bB  cC2Cl mol 2.1xNa mol 22Cl mol 1Na mol 4.2xNa mol 22Cl mol 1Na mol 4.2x-NaCl mol 4.2xNa mol 2NaCl mol 2Na mol 4.2xNa mol 2NaCl mol 2Na mol 4.2x-acC molA molbaB molA mol --John A. SchreifelsChem 211 10STOICHIOMETRY2•Mass of one reactant can be related to the mass of another reactant or product. (Law of definite proportions).E.g. Determine the mass of Na needed to react with 34.45 g of Cl2 and the maximum mass of NaCl that could be produced.–Write reaction and express the moles of one compared to the moles of the other. –Substitute for mol in terms of mass and formula mass in each part.Let n = mol; FM = formula mass thenE.g. 1: Determine the mass of oxygen consumed when it reacts with 10 g CH3CHO.E.g. 2. Calculate the mass of oxygen needed to react with 100 g Al to for Al2O3. 22NaCl mol122Cl molNa mol --FMmn John A. SchreifelsChem 211 11LIMITING REAGENTS•Limiting reagent: the reactant which limits the maximum amount of product that can be produced. It will be completely consumed in the reaction before any other reactants. •Limiting reagent must be known before theoretical yield can be determined. E.g. 1 Determine limiting reagent if 3.00 moles of Al react with 2.15 moles of O2 to form Al2O3.Strategy: –Determine mol of Al2O3 that could be produced from Al–Determine mol of Al2O3 that could be produced from O2–Reactant producing smallest amount of Al2O3 is limiting reagent. E.g.2 Calculate the theoretical yield when 20 g Al react with 25 g O2. Strategy: –Determine expected yield from each reactant.–Compare. If mass of Al2O3 from produced from Al is less than from O2, then Al is limiting. Otherwise O2 is limiting.John A. SchreifelsChem 211 12YIELDS IN CHEMICAL REACTIONS•Theoretical yield: the maximum amount of product that can be obtained from given amounts of reactants.•Actual Yield : the actual amount of product obtained from a reaction.•Theoretical yield calculated from amount of reactants and stoichiometry •Actual yield is the amount of product actually obtained.•% yield is the yield as a percent of the theoretical yield:•E.g. The reaction for the synthesis of acetic acid from methanol is shown below. What is the % yield if 15.0 g of methanol were combined with a stoichiometric amount of carbon monoxide to form19.1 g of